The random variable X can take values 0, 1, 2. If \(P(X = 0) = P(X = 1) = \alpha\), and \(E(X^2) = E(X)\), then which of the following are correct?
(A) \(E(X) = 2 - 3\alpha\)
(B) \(E(X^2) = 4 + 7\alpha\)
(C) \(\alpha = \displaystyle\frac{1}{2}\)
(D) \(\alpha = \displaystyle\frac{1}{5}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) and (C) only
The probabilities add up to 1, so
\(P(X = 2) = 1 - 2\alpha\)
Then
\(E(X) = 0(\alpha) + 1(\alpha) + 2(1 - 2\alpha)\)
\(E(X) = 2 - 3\alpha\)
\(E(X^2) = 0(\alpha) + 1(\alpha) + 4(1 - 2\alpha)\)
\(E(X^2) = 4 - 7\alpha\)
Using \(E(X^2) = E(X)\):
\(4 - 7\alpha = 2 - 3\alpha\)
\(4\alpha = 2\)
\( \displaystyle \alpha = \frac{1}{2} \)
(Then \(P(X = 2) = 0\), which is still a valid distribution.)
So \(E(X) = 2 - 3\alpha\), \(E(X^2) = 4 - 7\alpha\) and \(\alpha = \frac{1}{2}\).
Checking the statements
(A) \(E(X) = 2 - 3\alpha\) — Correct. \(\alpha + 2(1 - 2\alpha) = 2 - 3\alpha\).
(B) \(E(X^2) = 4 + 7\alpha\) — Incorrect. \(E(X^2) = 4 - 7\alpha\).
(C) \(\alpha = \frac{1}{2}\) — Correct. From \(4 - 7\alpha = 2 - 3\alpha\).
(D) \(\alpha = \frac{1}{5}\) — Incorrect. \(\alpha = \frac{1}{2}\).
NCERT Exemplar:
Probability
